Solve the equation. $\dfrac18 + c = \dfrac45$ $c=$
Let's subtract to get $c$ by itself. $\begin{aligned}\dfrac18 + c &= \dfrac45 \\ \\ \dfrac18 +c{-\dfrac18}&= \dfrac45{-\dfrac18}~~~~~{\text{subtract }\dfrac18} \text{ from each side to get } c \text{ by itself }\\ \\ \cancel{ \dfrac18} +c {{-}\cancel{{\dfrac18}}}&= \dfrac45{-\dfrac18}\\ \\ c &= \dfrac45{-\dfrac18}\end{aligned}$ $\begin{aligned} \dfrac45-\dfrac18 &=\dfrac{4\times8}{5\times8}-\dfrac{1\times5}{8\times5}\\\\ &= \dfrac{32}{40}-\dfrac{5}{40} \\\\ &= \dfrac{27}{40} \end{aligned}$ The answer: $c={\dfrac{27}{40}}$ Let's check to make sure. $\begin{aligned} \dfrac18 + c &= \dfrac45 \\\\ \dfrac18+{\dfrac{27}{40}}&\stackrel{?}{=} \dfrac45 \\\\ \dfrac5{40}+{\dfrac{27}{40}}&\stackrel{?}{=} \dfrac45 \\\\ \dfrac{32}{40}&\stackrel{?}{=} \dfrac45 \\\\ \dfrac45 &= \dfrac45 ~~~~~~~~~~\text{Yes!} \end{aligned}$